Angle properties of circles | Year 11

01 / LEARN

Understand the idea.

Identify the arc first

Before applying a circle angle theorem, identify its intercepted arc. An inscribed angle has its vertex on the circle and intercepts the arc between its chord endpoints that does not contain that vertex. The corresponding central angle may be reflex.

The central angle on that same arc is twice the inscribed angle. Angles subtended by one chord from the same segment are equal; vertices on opposite sides of the chord instead produce supplementary inscribed angles.

Use the diameter as a special case

A diameter cuts off a semicircle with central angle $180^{\circ}$ . Halving this gives a right angle at any other point on the circle. Thus a triangle drawn on a diameter has that diameter as its hypotenuse.

Join the centre to relevant vertices when building a proof: radii form isosceles triangles and connect their base angles to central angles. In a calculation, state the circle theorem before using right-triangle algebra; the sketch alone does not establish a right angle.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

A shared minor arc

The minor arc AB has central angle $132^{\circ}$ . P is on the major arc AB. Find $∠ A P B$ .

Worked solution

1
Identify the intercepted arc.
Because P lies on the major arc AB, angle APB intercepts the minor arc AB.
2
Apply the central-angle theorem.
$2 ∠ A P B = 132^{\circ}$
The central angle on the same minor arc is twice the inscribed angle.
3
Halve the central angle.
$∠ A P B = 66^{\circ}$
Dividing the established angle relation by two gives the required angle.

Answer

$66^{\circ}$ .

EXAMPLE 02

A diameter gives a length

AB is a diameter and P is on the circle. If $A P = 10$ cm and $B P = 24$ cm, find AB.

Worked solution

1
Establish the right angle.
$∠ A P B = 90^{\circ}$
An angle subtended by a diameter at the circumference is a right angle.
2
Apply Pythagoras with AB as hypotenuse.
$\begin{aligned} A B^{2} & = A P^{2} + B P^{2} \\ = 10^{2} + 24^{2} = 676 \end{aligned}$
The two given chords are the perpendicular shorter sides of triangle APB.
3
Choose the positive length.
$A B = \sqrt{676} = 26 cm$
A segment length is positive, so the negative algebraic root is inadmissible.

Answer

$A B = 26$ cm.

03 / YOUR TURN

Try it yourself.

Check your understanding

The minor central angle AOB is $100^{\circ}$ and P is on the major arc AB. Find angle APB.

Show answer & reasoning

$50^{\circ}$ .

P lies on the major arc, so angle APB intercepts the minor arc. The inscribed angle is $100^{\circ} \div 2 = 50^{\circ}$ .

Keep practising

6 worksheet questions · 14 marks

Open worksheet Worked solutions

Read the questions online

An inscribed angle subtends the same arc as a central angle of $146^{\circ}$ . Find it.
2 marks
An inscribed angle is $48^{\circ}$ . Find the central angle on its intercepted arc.
2 marks
$P$ and $Q$ lie on the same segment of chord $A B$ . If $∠ A P B = 57^{\circ}$ , find $∠ A Q B$ .
2 marks
$A B$ is a diameter. In triangle $A P B$ , angle $P A B$ is $31^{\circ}$ . Find angle $P B A$ .
2 marks
The major arc $A B$ has central reflex angle $248^{\circ}$ . $P$ lies on the minor arc $A B$ . Find angle $A P B$ .
3 marks
Explain why two angles subtending chord $A B$ from opposite segments sum to $180^{\circ}$ .
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

An inscribed angle subtends the same arc as a central angle of $146^{\circ}$ . Find it.
2 marks
$\frac{146}{2} = 73$ .
$73^{\circ}$ .
An inscribed angle is $48^{\circ}$ . Find the central angle on its intercepted arc.
2 marks
The central angle is twice the inscribed angle.
$96^{\circ}$ .
$P$ and $Q$ lie on the same segment of chord $A B$ . If $∠ A P B = 57^{\circ}$ , find $∠ A Q B$ .
2 marks
Angles in the same segment are equal.
$57^{\circ}$ .
$A B$ is a diameter. In triangle $A P B$ , angle $P A B$ is $31^{\circ}$ . Find angle $P B A$ .
2 marks
Angle $A P B$ is $90^{\circ}$ ; use $180 - 90 - 31 = 59$ .
$59^{\circ}$ .
The major arc $A B$ has central reflex angle $248^{\circ}$ . $P$ lies on the minor arc $A B$ . Find angle $A P B$ .
3 marks
$P$ intercepts the major arc, so the correct calculation is $\frac{248}{2} = 124$ .
$124^{\circ}$ .
Explain why two angles subtending chord $A B$ from opposite segments sum to $180^{\circ}$ .
3 marks
Each inscribed angle is half its intercepted arc measure.
The two arcs make the complete circle, so the angle sum is $\frac{360^{\circ}}{2} = 180^{\circ}$ .
They intercept arcs whose measures total $360^{\circ}$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 Mathematics: geometric reasoning and real-number/proof material, regrouped into a school chapter; all exercises are original.

Official curriculum source ↗

In this lesson

Understand the idea.

Identify the arc first

Use the diameter as a special case

See the method in action.

A shared minor arc

Identify the intercepted arc.

Apply the central-angle theorem.

Halve the central angle.

A diameter gives a length

Establish the right angle.

Apply Pythagoras with AB as hypotenuse.

Choose the positive length.

Try it yourself.

Check your understanding

Keep practising